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17 August, 13:03

Some fertilizer blends contain magnesium nitrate (MgNO3). Suppose that a chemist has 1.24 liters of a 2.13 M solution of magnesium nitrate. If the chemist dilutes the solution to 1.60 M, what is the volume of the new solution?

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  1. 17 August, 14:22
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    Initial concentration of magnesium nitrate M1 = 2.13 M

    Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L

    Final concentration of MgNO3, M2 = 1.60 M

    Let the final volume of MgNO3 upon dilution be V2

    Formula to use:

    M1*V1 = M2*V2

    V2 = M1*V1/M2

    = 2.13 M * 1.24 L/1.60 M = 1.65 L

    Thus, the final volume of magnesium nitrate solution upon dilution is 1.65 L
  2. 17 August, 15:35
    0
    Initial concentration of magnesium nitrate M1 = 2.13 M

    Initial volume of magnesium nitrate, MgNO3 V1 = 1.24 L

    Final concentration of MgNO3, M2 = 1.60 M

    Let the final volume of MgNO3 upon dilution be V2

    Formula to use:

    M1*V1 = M2*V2

    V2 = M1*V1/M2

    = 2.13 M * 1.24 L/1.60 M = 1.65 L

    I believe I did it right ...
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